Problem: Here's a parameterization of a plane: $\vec{v}(x, y) = (2y - x, 4x + 3y, x - y)$ What vectors are normal to the plane $\vec{v}$ ? Choose 2 answers: Choose 2 answers: (Choice A) A $(-7, 1, -11)$ (Choice B) B $(7, -1, 11)$ (Choice C) C $(-11, 1, 7)$ (Choice D) D $(11, -1, -7)$
Solution: The vector normal to the area element describes what's perpendicular to the surface, and the vector's magnitude represents the area of a tiny rectangle along the parameterization. We can take multiply the result by $-1$ to find another normal vector. $\text{normal to area element} = \dfrac{\partial \vec{v}}{\partial x} \times \dfrac{\partial \vec{v}}{\partial y}$ Let's calculate the cross product. $\begin{aligned} \dfrac{\partial \vec{v}}{\partial x} \times \dfrac{\partial \vec{v}}{\partial y} &= \det \begin{pmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \\ -1 & 4 & 1 \\ \\ 2 & 3 & -1 \end{pmatrix} \\ \\ &= -7 \hat{\imath} + \hat{\jmath} - 11 \hat{k} \end{aligned}$ We want two normal vectors. Because the negative of a normal vector is also normal to the surface, we can take the negative of what we just calculated as a second normal vector to the plane $\vec{v}$. Therefore, two vectors normal to $\vec{v}$ are $(-7, 1, -11)$ and $(7, -1, 11)$.